# Show that if X and Y are independent N(0, 1)-distributed random variables, then X/Y ∈ C(0, 1).

Show that if $X$ and $Y$ are independent $N(0, 1)$-distributed random variables, then $X/Y ∈ C(0, 1)$. C is cauchy distribution..

First trial: 1) Transformation

Let $Y=V,U=X/V,=>X=UV$;$|J|=v$

$f(u,v)=f_x(uv)f_y(v)|J|=1/2\pi ve^{-\frac{(uv)^2}{2}}e^{-\frac{v^2}2}$

To get $f_u(u) =\int_{-∞}^{+∞}1/2\pi ve^{-\frac{(uv)^2}{2}}e^{-\frac{v^2}2}=0$

Second trial: 2) CDF

$F(U\le u)=F(X \le uY)=\int_{-∞}^{+∞}F(X \le uy)f_y(y)dy=1/2\pi\int_{-∞}^{+∞} (\int_{-∞}^{uy}e^{-\frac{(uy)^2}{2}}du)e^{-\frac{y^2}2}dy=?$

Can you help me?

I carry on your second trial: $$\frac{X}Y\leq u$$

Now we discriminate between two cases $$Y>0$$ and $$Y<0$$. Multiplying the equation above by $$Y$$.

$$P(\frac{X}Y\leq u)=(X\leq uY| Y>0)+P(X\geq uY| Y<0)$$

$$=\int_{0}^{\infty}\int_{0}^{uy} f(x,y) \,dx \, dy +\int_{-\infty}^{0}\int_{uy}^{0} \,dx \, dy$$

Then we differentiate w.r.t. u to get the pdf. This can be done by using the Leibniz rule.

$$p(u)=\int_{0}^{\infty} y\cdot f(uy,y) \, dy- \int_{-\infty}^{0} y\cdot f(uy,y) \, dy$$

$$=2\int_{0}^{\infty} y\cdot f(uy,y) \, dy$$

$$=2\cdot\frac{1}{2\pi\sigma_x\sigma_y}\int_{0}^{\infty} y\cdot \exp\left( -\left(\frac{y^2}{2\sigma_x^2}+\frac{u^2y^2}{2\sigma_y^2}\right)\right) \, dy$$

$$=\frac{1}{\pi\sigma_x\sigma_y}\int_{0}^{\infty} y\cdot \exp\left( -y^2\left(\frac{1}{2\sigma_x^2}+\frac{u^2}{2\sigma_y^2}\right)\right) \, dy$$

Let $$c=\frac{1}{2\sigma_x^2}+\frac{u^2}{2\sigma_y^2}$$. The integral becomes equal to

$$\int_{0}^{\infty} y\cdot \exp\left( -cy^2\right) \, dy$$, which is $$\frac1{2c}$$. Thus we have

$$p(u)=\frac{1}{\pi\sigma_x\sigma_y}\cdot \frac{1}{2\cdot(\frac{1}{ 2\sigma_x^2}+\frac{u^2}{2\sigma_y^2})}=\frac{1}{\pi\sigma_x\sigma_y}\cdot \frac{1}{\frac{1}{ \sigma_x^2}+\frac{u^2}{\sigma_y^2}}$$

$$=\boxed{\large{\frac1{\pi}\frac{\frac{\sigma_y}{\sigma_x}}{u^2+\left(\frac{\sigma_y}{\sigma_x}\right)^2}}}$$

This is the pdf of the cauchy distribution with $$\gamma=\frac{\sigma_y}{\sigma_x}$$ and $$x_0=0$$.